# Proving Cauchy-Schwarz with Arithmetic Geometric mean

I understand common proofs for Cauchy-Schwarz, but not sure about the first step in this one, which is proof 4 from here

Let $$A = \sqrt{a_1^2 + a_2^2 + \dots + a_n^2}$$ and $$B = \sqrt{b_1^2 + b_2^2 + \dots + b_n^2}$$. By the arithmetic-geometric means inequality (AGI), we have

$$\sum_{i=1}^n \frac{a_ib_i}{AB} \leq \sum_{i=1}^n \frac{1}{2} \left( \frac{a_i^2}{A^2} + \frac{b_i^2}{B^2} \right) = 1$$

so that

$$\sum_{i=1}^na_ib_i \leq AB \leq \sqrt{\sum_{i=1}^na_i^2} \sqrt{\sum_{i=1}^n b_i^2}$$

Which is Cauchy-Schwarz. Now this is all quite elegant, but how is the first equation RHS equal to 1? And how is it AGI? Shouldn't it be in this form:

$$\sqrt[n]{\prod_{i=1}^n x_i} \leq \frac{1}{n} \sum_{i=1}^n x_i$$

There's probably something simple I'm missing...

• – glS Aug 26 at 11:43

First, $$\sum_{i=1}^{n} \frac{a_i^2}{A^2} = \frac{1}{A^2}\sum_{i=1}^{n} a_i^2 = \frac{A^2}{A^2} = 1.$$ Similar could be said about the second term in the summation.
Second, the proof uses the following two-variable AM-GM inequality: $$xy = \sqrt{x^2 y^2} \le \frac{x^2+y^2}{2}.$$
• except there is also the trivial inequality $x\leq |x|$ since $a_i,b_i$ are not assumed to be nonegative. – user10354138 Sep 21 '18 at 18:46
• @vega $x=\frac{a_i}{A}$ and $y = \frac{b_i}{B}$. – Math Lover Sep 21 '18 at 21:34